What is the solution for y^{''}+4y=0,y(3pi)=-3,y(0)=-3 ?

The solution for y^{''}+4y=0,y(3pi)=-3,y(0)=-3 is y=-3cos(2t)+c_{2}sin(2t)

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y^{''}+4y=0,y(3pi)=-3,y(0)=-3

y^{^{''}} + 4 y = 0, y (3 π) = - 3, y (0) = - 3

y = - 3 cos (2 t) + c_{2} sin (2 t)

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

y = - 3 cos (2 t) + c_{2} sin (2 t)

y = - 3 cos (2 t) + c_{2} sin (2 t)

y^{^{''}} - y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 1

2 y^{^{''}} + 14 y^{^{'}} + 12 y = 0

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 2

What is the solution for y^{''}+4y=0,y(3pi)=-3,y(0)=-3 ?
The solution for y^{''}+4y=0,y(3pi)=-3,y(0)=-3 is y=-3cos(2t)+c_{2}sin(2t)